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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 751. |
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. what will be length of the air column above mercury in the above now ? (Atmospheric pressure = 76 cm of Hg)A. 22 cmB. 38cmC. 6cmD. 16cm |
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Answer» Correct Answer - D `76xx8=(76+x-54)x` |
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| 752. |
An ideal gas A and a real gas B have their volumes increases from `V to 2V` under isothermal condtitions. The increase in internal energyA. will be same in both A and BB. will be zero in bothe the gasesC. of B will be more than that of AD. of A will be more than that of B |
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Answer» Correct Answer - B `Delta U = Delra Q - p DeltaV = DeltaQ-pV`. |
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| 753. |
If pressure of `CO_(2)` (real gas ) in a container is given by `P = (RT)/(2V - b) - (a)/(4b^(2))`, then mass of the gas in container isA. `11 g`B. `22 g`C. `33 g`D. `44 g` |
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Answer» Correct Answer - C van der Walls gas equation for `n` moles of real gas `(P + (n^(2)a)/(V^(2))) (V - nb) = nRT` `P = (nRT)/(V - nb) - (n^(2)a)/(V^(2)) = (RT)/((V)/(n) - b) - (a)/(V^(2)//n^(2))` (i) Given equation `P = (RT)/(2V - b) - (a)/(4b^(2))` (ii) Comparing (i) and (ii) `n = (1)/(2) = (M)/(M_(0)) = (M)/(44)` `M = 22 g` |
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| 754. |
The equation of state of gas is given `(P + (aT^(2))/(V))V^( c) = (RT + b)` where `a, b, c` and `R` are constant. The isotherms can be represented by `P = AV^(m) - BV^(n)`, where `A` and `B` depend only on temperature andA. `m = -c` and `n = -1`B. `m = c` and `n = 1`C. `m = - c` and `n = 1`D. `m = c` and `n = -1` |
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Answer» Correct Answer - A `(P + (aT^(2))/(V)) V^(c ) = RT + b` `P = (RT + b)/(V^(c )) - (aT^(2))/(V)` (i) `P = AV^(m) - BV^(n)` (ii) Comparing (i) and (ii) `m = - c, n = -1` |
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| 755. |
Fig shows of `PV//T` versus P for `1.00 xx 10^(-3) kg` of oxygen gas at two different temperatures. (a) What does the dotted plot signify ? (b) Which is true : `T_(1) lt T_(2) or T_(2) lt T_(1) ?` ( c) What is the value of `PV//T` where the curves meet on the Y-axis ? (d) If we obtained similar plot for `1.00 xx 10^(-3) kg` of hydrogen, would we get the same value of `PV//T` at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yield the same value of `PV//T` (for low pressure high temperature region of the plot) ? (Molecular mass of `H = 2.02 u`, of `O = 32.0 u, R = 8.31 J "mol"^(-1) K^(-1)` |
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Answer» (i) For the dotted line, `(PV)/(T)=` constant. It represents the behaviour of an ideal gas (ii) The gas equation PV=RT is true only at high temperature. In figure, curve for `T_(1)` is coloser to the dotted line than curve for `T_(2)` Hence the relation between `T_(1)` and `T_(2)` is `T_(1)gtT_(2)` (iii) `(PV)/(T)=muR` , Here `mu=(m)/(M)=(2)/(32)=(1)/(16)` `therefore (PV)/(T)=(1)/(16)xx8.31=0.519 J//K` |
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| 756. |
`A` cylinder of fixed capacity `67.2` liters contains helium gas at `STP` . Calculate the amount of heat required to rise the temperature of the gas by `15^(@)C` ? `(R=8.314 J mol^(-1)k^(-1))`A. 520JB. 560.9JC. 620JD. 621.2J |
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Answer» Correct Answer - B `dQ=nC_(v)dT` |
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| 757. |
A cylinder of fixed capacity 22.4 litre contains helium gas at standard temperature and pressure. What is the amount of heavy needed to raise the temperature of the gas in the cylinder by `30^(@)C` ? `(R=8.31J "mol"^(-1)K^(-1))` |
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Answer» 1 mole of any gas occupies 22.4 litres. Since the volume of cylinder is fixed, the heat required `nC_(v)dT` Here Helium is monoatomic gas so `C_(v)=(3)/(2)K` Heat required `=1xx1.5Rxx30=45R=374.5J` |
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| 758. |
A cylinder of fixed capacity (of 44.8 L) contains 2 moles of helium gas at STP .What is the amount of heat needed to raise the temperature of the gas in the cylinder by `20^(@)C` ? `("Use"R=8.31J"mol"^(-1)K^(-1))`A. 996 JB. 831 JC. 498 JD. 374 J |
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Answer» Correct Answer - C Since , the volume of cylinder is fixed and helium is a monoatomic gas. The internal energy = `muCvDeltaT` `U=(1)/(2)fmuRDeltaT" " [thereforeC_(v)=(1)/(2)fR]` `U=(1)/(2)xx3xx2xx8.31xx20=20=498.6J` Carnot Engine and Refrigerator |
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| 759. |
If the density of oxygen is 1.44 `kg//m^(3)` at pressure of `10^(5)N//m^(2)`, then the root-mean-square velocity of oxygen molecules in m/s will beA. 469B. 456C. 120D. 270 |
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Answer» Correct Answer - B `C=sqrt((3P)/(rho))=sqrt((3xx10^(5))/(1.44))=(1.732xx3.13xx10^2)/(1.2)` `=456m//s`. |
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| 760. |
`(1//2)` mole of helium is contained in a container at STP how much heat energy is needed to double the pressure of the gas, keeping the volume constant? Heat capacity of gas is `3 J g^(-1)K^(-1)`.A. 3276 JB. 1638JC. 819JD. 409.5J |
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Answer» Correct Answer - B `(dQ)_(v)=nC_(v)dT` |
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| 761. |
The root mean square velocity of gas molecules of `0^(@)C` will be if at N.T.P. its density is 1.43 kg/`m^(3)`A. 461m/sB. 164m/sC. 461cm/sD. 164cm/s |
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Answer» Correct Answer - A `v_(rms)=sqrt((3P)/(rho))` |
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| 762. |
For a gas `R/C_V` = 0.4, where R is the universal gas constant and C, is molar specific heat at constant volume. The gas is made up of molecules which areA. rigid diatomicB. monoatomicC. non-rigid diatomicD. polyatomic |
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Answer» Correct Answer - A According to question , `(R)/(C_(v))=0.4` [R = universal constant and `C_(v)`=molar specific heat at constant volume] As we know, `C_(p)-C_(v)=R` `"So " (C_(p)C_(v))/(C_(v))=0.4` `rArr" " (C_(p))/(C_(v))=0.4+1=1.4` `"i.e "Y=1.4` As Y = 1.4 , the gas is diatomic in nature. |
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| 763. |
The molar specific heat of an ideal gas at constant pressure and constant volume is `C_(p)` and `C_(v)` respectively. If R is the universal gas constant and the ratio of `C_(p)` to `C_(v)` is `gamma`, then `C_(v)`.A. `(1+gamma)/(1-gamma)`B. `(R)/((gamma-1))`C. `((gamma-1))/(R)`D. `gammaR` |
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Answer» Correct Answer - B The value of `C_(v)=(R)/(gamma-1)` |
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| 764. |
If universal gas constant is R, the essential heat to increase from K to 473 K at constant volume for ideal gas of 4 moles isA. 200 RB. 400 RC. 800 RD. 1200 R |
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Answer» Correct Answer - D Specific heat for a monoatomic gas `C_(v)=(3)/(2)R` `therefore"Heat "dQ=muC_(v)DeltaT` `dQ=muxx(3)/(2)xxR(473-273)` = `4xx(3)/(2)xxRxx200" " (becausemu=4)` `therefore" " dQ=4xx300R=1200R` |
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| 765. |
The value of universal gas constant is 8.3J/mole-k, The mean kinetic energy of 32gm of oxygen at `.-73^(@)C` will beA. 480JB. 4980JC. 2490JD. 100J |
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Answer» Correct Answer - C `E=(3)/(2)(m)/(M)RT` |
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| 766. |
The temperature of sink of Carnot engine is `27^(@)C`. Efficiency of engine is `25%`. Then temeperature of source isA. `27^(@)C`B. `127^(@)C`C. `327^(@)C`D. `227^(@)C` |
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Answer» Correct Answer - B `eta=1-(T_(2))/(T_(1))=(1)/(4)=1-(300)/(T_(1)),(300)/(T_(1))=(3)/(4):` `T_(1)=400K` `t_(1)=400-273` `=127^(@)C`. |
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| 767. |
The efficiency of a Carnot engine operating between temperatures of `100^(@)C` and `-23^(@)C` will beA. `(100-23)/(100)`B. `(100-23)/(373)`C. `(100+23)/(373)`D. `(100+23)/(100)` |
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Answer» Correct Answer - C `eta=(T_(1)-T_(2))/(T_(1))=((273+100)-(273-23))/(273+100)=(100+23)/(373)` |
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| 768. |
P-V graph for an ideal gas undergoing polytropic process `PV^(m) = constant` is shown here. Find the value of m. A. `3/4`B. `-3/2`C. `5/3`D. `3/2` |
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Answer» Correct Answer - D `PV^(m) = const`. `implies V^(m) dP + mv^(m-1) Pd v=0` `implies (dP)/(dV) = -(mP)/(V) = tan(180 - 37^@)` `=3/4 = m(2xx10^5)/(4xx10^5) implies m = 3/2`. |
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| 769. |
A polytropic process for an ideal gas is represented by equation `PV^(n) = constant`. If g is ratio of specific heats `((C_(p))/(C_(v)))`, then value of n for which molar heat capacity of the process is negative is given asA. `gamma gt n`B. `gamma gt n gt1`C. `n gt gamma`D. none, as it is not possible |
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Answer» Correct Answer - B `C = R/(gamma-1) - R/(n-1) = (R(n-gamma))/((n-1)(gamma-1))` `C` is negative if `gamma gt n gt 1`. |
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| 770. |
A hypothetical gas sample has its molecular speed distribution graph as shown in the figure. The speed (u) and `(dN)/(du)` have appropriate units. Find the root mean square speed of the molecules. Do not worry about units. |
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Answer» Correct Answer - `mu_(rms)=sqrt((8)/(3))` |
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| 771. |
A sample of an ideal gas initially having internal energy `U_(1)` is allowed to expand adiabatically performing work W. Heat Q is then supplied to it, keeping the volume constant at its new value, until the pressure raised to its original value. The internal energy is then `U_(3)` (see figure). find the increase in internal energy `(U_(3) - U_(1))`? A. `Q + W`B. `Q - W`C. `gamma W - Q`D. `Q - gamma W` |
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Answer» Correct Answer - B Process `1 rarr 2 ` is adiabatic `Q_(12) = DeltaU_(12) + W_(12) = gt 0 = (U_2-U_1) + W` `U_2 - U_1 = -W` Process `2 rarr 3` is isochoric `Q_(23) = Delta U_(23) + W_(23) implies Q = U_(3)-U_(2)` `W_(23) = 0 :. V = consta nt` `implies U_(3) -U_(2) = Q` Now `(U_(3)-U_(2)) + (U_(2)-U_(1)) = U_(3)-U_(1)` `implies Q _(-W) = U_(3) - U_(1)` `implies U_3 = Q +U_(1) -W implies U_(3) -U_(1) = Q-W`. |
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| 772. |
A certain mass of a gas is compressed first adiabatically, and then isothermally. In both cases, the initial state of the gas is the same. Is the work done `W_(1)` in the first case greater than the work done `W_(2)` in the second case ? Explain. |
| Answer» Yes, because the area between curves 1 and trhe V-axis is greater than that between curve 2 and the V-axis and the work done is given by the area between the `p - V` curve and the V-axis. | |
| 773. |
While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a poistion where it can stay in equilibrium. In this condition, distance of the piston from the top isA. `(2P_(0)phi^(2))/(phir^(2)P_(0)+Mg)(2L)`B. `(P_(0)phiR_(2)-Mg)/(phiR^(2)P_(0))(2L)`C. `(P_(0)phiR^(2)+Mg)/(phir_(2)P_(0))(2L)`D. `(P_(0)phiR^(2))/(phiR^(2P_(0)-Mg))(2L)` |
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Answer» Correct Answer - C |
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| 774. |
The change in internal energy of a gas kept in a rigid cylinder on supplying 120 J of heat isA. 0 JB. 60 JC. 100 JD. 120 J |
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Answer» Correct Answer - D `DeltaU=DeltaQ` =120 J (`becauseDeltaw=0`) |
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| 775. |
One mole of a monoatomic ideal gas undergoes the process `ArarrB` in the given P-V diagram. What is the specific heat for this process? A. `(P_0V_0)/(R)`B. `(3P_0V_0)/(R)`C. `(9P_0V_0)/(8R)`D. None of these |
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Answer» Correct Answer - C max. at mid point `((3V_0)/(2),(3P_0)/(4)) rArr T = (9)/(8) (P_(0)V_(0))/(2)` |
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| 776. |
One mole of a monoatomic ideal gas undergoes the process `ArarrB` in the given P-V diagram. What is the specific heat for this process? |
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Answer» Specific heat `C=(DeltaQ)/(DeltaT)=(1)/(DeltaT)(DeltaU+W)=C,+(W)/(DeltaT)` For the given process `W=4V_(0)(9P_(0))/(2)=18P_(0)V_(0) (because W="area of" P-V "graph")` Also, `DeltaT=T_(2)-T_(1)` `((6P_(0))(5V_(0)))/(R)-((3P_(0))V_(0))/(R)=(27P_(0)V_(0))/(R)` and `C_(v)=(3)/(2)R` `therefore C=C_(v)+(W)/(DeltaT)=(3R)/(2)+(18P_(0)V_(0))/(((27P_(0)V_(0))/(R)))` `=(3R)/(2)+(2R)/(3)=(13R)/(6)` |
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| 777. |
A balloon containing an ideal gas has a volume of `10 litre` and temperature of `17^(@)`. If it is heated slowly to `75^(@)C`, the work done by the gas inside the balloon is (neglect elasticity of the balloon and take atmospheric pressure as `10^(5)` Pa)A. `100 J`B. `200 J`C. `250 J`D. data insufficient |
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Answer» Correct Answer - B Since elasticity of balloon is negligible, pressure inside balloon pressure outside balloon = `P_(atm)` `:. W = P_(atm) DeltaV` `V_(i n) = 10 litre` `(V_(i n))/(T_(i n)) = (V_(fi n))/(T_(fi n) ) implies V_(fi nal) = ((V_(i n)T_(fi nal))/(T_(i n))) litre` `implies W = P_(atm) V_(i n) ((T_(f i nal))/(T_(i n))-1)` ltbr gt`implies 10^(5)xx10^(-2)(58/290)= 200J`. |
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| 778. |
The ratio of the specific heats `(C_(P))/(C_(upsilon)) = gamma` in terms of degrees of freedom is given byA. `(1+(1)/(n))`B. `(1+(n)/(3))`C. `(1+(2)/(n))`D. `(1+(n)/(2))` |
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Answer» Correct Answer - C The ratio of the specific heats in relation with Degrees of freedom is given by `gamma=1+2/(n)` |
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| 779. |
Value of two principal specific heats of a gas `("in cal mol" K)^(-1)` determined by different students are given . Which is most relible ?A. 5,2B. 6,5C. 7,5D. 7,4 |
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Answer» Correct Answer - C As, `C_(p)-C_(v)=R=2"cal"("mol"K)^(-1)` . Difference in the two values must be 2. `C_(p)=7and C_(v)=5` |
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| 780. |
Each molecule of a gas has F degrees of freedom . The ratio `(C_(p))/(C_(V))=gamma` for the gas isA. `(2)/(f)+1`B. `1-(2)/(f)`C. `1+(1)/(f)`D. `1-(1)/(f)` |
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Answer» Correct Answer - A As , `eta=(C_(p))/(C_(v))=1+(2)/(n)=1+(2)/(f)` |
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| 781. |
Caluculate Change in internal energy when 5 moles of hydrogen is heated to `20^(@)C` from , Specific heat of hydrogen at constant pressure is 8 cal `("mol"^(@)C)^(-1)`A. 200 calB. 350 calC. 300 calD. 450 cal |
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Answer» Correct Answer - C Given , `C_(p)=8 "cal"("mol"^(@)C)^(-1)` `C_(v)=C_(p)-R=8-2=6" cal "("mol"^(@)C)` `therefore"internal energy du"=mC_(v)(T_(2)-T_(1))=5xx6(20-10)` `300"cal" ` |
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| 782. |
A gas is found to be obeyed the law `p^2V = constant`. The initial temperature and volume are `T_0 and V_0`. If the gas expands to a volume `3 V_0`, then the final temperature becomes.A. `T/(sqrt(3))`B. 3TC. `T/3`D. `Tsqrt(3)` |
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Answer» Correct Answer - D From equation PV=nRT and `P^(2) V=` constant `(T^(2))/V`=constant `T_(2)=sqrt(3)T` |
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| 783. |
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increase as `V^q`, where V is the volume of the gas. The value of q is : `(gamma=(C_p)/(C_v))`A. `(3gamma+5)/(6)`B. `(3gamma-5)/(6)`C. `(gamma+1)/(2)`D. `(gamma-1)/(2)` |
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Answer» Correct Answer - C `tau=(gamma)/(V_("max"))=(1)/(sqrt(2)pid^(2)((N)/(V))sqrt((3RT)/(M)))` `tauprop(V)/sqrt(T)` `TV^(gamma-1)=k` `taupropV^((gamma+1)/(2))` |
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| 784. |
Assertion: In an isolated system the entropy increases. Reason: The processes in an isolated system are adiabatic.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false |
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Answer» Correct Answer - B The entropy change of the universe during reversible processes is zero. During an adiabatic process no heat flows. Thus, we see that the entropy change of a systen during a reversible, adiabatic process is zero. But note that boht qualifers are needed, the entropy of a non-isolated system change during a reversible process (and the entropy change of the surrounding will compensate), and an irreversible change to an isolated system will increase the entrophy. hence option (b) is true. |
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| 785. |
If an electric fan be switched on in a closed room, will the air of the room be cooled? If not, why do we feel cold?A. cooledB. heatedC. maintains its temperatureD. heated or cooled depending on the atmospheric pressure |
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Answer» Correct Answer - B When the electric fan is switched on, the electric energy is converted into mechanical enerty, which is turen is converted into heat energy. It is so because the rotainty fan in an a closed room will increases the kinetic. Energy of translation of molecules of air in the room and hence, temperature of the room increases. |
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| 786. |
If for hydrogen `C_(P) - C_(V) = m` and for nitrogen `C_(P) - C_(V) = n`, where `C_(P)` and `C_(V)` refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between `m` and `n` is (molecular weight of hydrogen = 2 and molecular weight or nitrogen = 14)A. a=16bB. b=16aC. a=4bD. a=b |
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Answer» Correct Answer - D Both are diatomic gases and `C_(P)-C_(V)=R` for all gases |
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| 787. |
If for hydrogen `C_(P) - C_(V) = m` and for nitrogen `C_(P) - C_(V) = n`, where `C_(P)` and `C_(V)` refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between `m` and `n` is (molecular weight of hydrogen = 2 and molecular weight or nitrogen = 14)A. `n = 14 m`B. `n = 7m`C. `m = 7n`D. `m = 14 n` |
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Answer» Correct Answer - C c.`C_(P) - C_(v) = m`, for hydrogen `(M_(1) = 2)` `C_(P) - C_(v) = n`, for nitrogen `(M_(2) = 14)` For hydrogen, `C_(P) - C_(v) = (1)/(M_(1)) (dQ)/(dT) = m` For nitrogen, `C_(P) - C_(v) = (1)/(M_(2)) (dQ)/(dT) = n` `:. (m)/(n) = ((1)/(M_(1)) (dQ)/(dT))/((1)/(M_(2)) (dQ)/(dT)) = (M_(2))/(M_(1)) = (14)/(2) = 7` `:. m = 7n` |
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| 788. |
`DeltaU=0` in a noncylic process of an ideal gas. The processA. may be isothermalB. must be isothermalC. may be adiabaticD. may be isobaric |
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Answer» Correct Answer - A Let A and B are two points on an isothermal curve. Join A and B by any curve C. Then on this curve C, `DeltaU=0` where c is not an isothermal curve. |
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| 789. |
In an adiabatic expansion the product of pressure and volume :A. decreasesB. increasesC. remains constantD. first increases then decreases. |
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Answer» Correct Answer - A In an adiabatic expansion, internal decreases and hence temperature decreases. `:.` from equation of state of ideal gas PV=nRT `rArr ` the product of P and V decreases |
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| 790. |
The `P - V` diagram of a system undergoing thermodynamic transformation is shown in figure. The work done by the system in going from `A rarr B rarr C is 30 J` and `40 J` heat is given to the system. The change in internal energy between `A and C` is A. `10 J`B. `70 J`C. `84 J`D. `134 J` |
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Answer» Correct Answer - A Heat given `DeltaQ =40J` and work done `DeltaW = 30J` `implies DeltaQ=DeltaQ-DeltaW = 40-30 = 10J`. |
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| 791. |
The `P - V` diagram of a system undergoing thermodynamic transformation is shown in figure. The work done by the system in going from `A rarr B rarr C is 30 J` and `40 J` heat is given to the system. The change in internal energy between `A and C` is A. 50 JB. 60 JC. 30 JD. 10 J |
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Answer» Correct Answer - A `U_(A) =0, U_(B) =20 U_(C) =?` `triangleQ_(BC) = 30` `triangleQ_(C) = triangleU_(BC) =U_(C)-U_(B) = 30` `U_(C) =U_(B) + 30 =50` |
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| 792. |
Pick the correct statement (s) :A. The rms translational speed for all ideal gas molecules at the same temperature is not the same but it depends on the massB. Each particle in a gas has average transloational kinetic energy and the equation `1//2 mv_(max)^(2) = 3//2 kT` establisehes the relationship between the average translational kinetic energy per particle and temperature of an ideal gas.C. If the temperature of an ideal gas is doubled form `100^(@)C` to `200^(@)C`, the average kinetic energy of each particle is also doubled.D. It is possible for both pressure and volume of a monatomic ideal gas to change simultaneously without casuing the internal enregy of the gas to change |
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Answer» Correct Answer - A::D `V_(rms)=sqrt((3kT)/( m))` Siince `PV=Nrt`, therefore `P` and `V` both can change simultaneously keeping the temperture constant. |
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| 793. |
An ideal gas undergoes an expansion from a state with temperature `T_(1)` and volume `V_(1)` through three different polytropic processes `A, B` and `C` as shown in the `P - V` diagram. If `|Delta E_(A)|, |Delta E_(B)|` and `|Delta E_(C )|` be the magnitude of changes in internal energy along the three paths respectively, then : A. `|Delta E_(A)|lt|Delta E_(B)|lt|Delta E_(C )|` if temperature in every process decreasesB. `|Delta E_(A)|gt|Delta E_(B)|gt|Delta E_(C )|` if temperature in every process decreasesC. `|Delta E_(A)|gt|Delta E_(B)|gt|Delta E_(C )|` if temperature in every process increasesD. `|Delta E_(B)|lt|Delta E_(A)|lt|Delta E_(C )|` if temperature in every process increases |
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Answer» Correct Answer - A::C Initial state is same for all the three processes (say initial internal energy `=E_(0))`. In the final state, `V_(A)=V_(B)=V_(C)` and `P_(A) gt P_(B) gt P_(C)` `P_(A)V_(A) gt P_(B)V_(B) gt P_(C)V_(C)` ` E_(A) gt E_(B) gt E_(C)` If `T_(1) lt T_(2)`, then `E_(0) gt E_(f)` for all the three processes and hence `(E_(0)-E_(A)) lt (E_(0)-E_(B)) lt (E_(0)-E_(C))` `|DeltaE_(A)|lt|DeltaE_(B)|lt|DeltaE_(C)|` If `T_(1)lt T_(2)`, then ` E_(0) lt E_(f)` for all three processes and hence `(E_(A)-E_(0)) gt (E_(B)-E_(0)) gt (E_(C)-E_(0))` `|DeltaE_(A)|gt|DeltaE_(B)|gt|DeltaE_(C)|` |
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| 794. |
Two vessel separately contains two ideal gases A and B at the same temperature, the pressure of A being twice that of B. under such conditions, the density of A is found to be 1.5 times the density of B. the ratio of molecular weight of A and B isA. `1/2`B. `2/3`C. `3/4`D. `2` |
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Answer» Correct Answer - C `PV=nRTimplies P =(rho)/(M)RT` `M = (rhoRT)/(P) implies M prop (rho)/(P)` `(M_1)/(M_2)=3/2 xx 1/2 implies M_(1):M_(2)=3:4`. |
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| 795. |
A system goes from A and B via two processes. I and II as shown in figure. If `DeltaU_1 and DeltaU_2` are the changes in internal energies in the processes I and II respectively, then A. `DeltaU_(||) lt DeltaU_(|)`B. `DeltaU_(||) lt DeltaU_(|)`C. `Delta U_(||)=Delta U_(|)`D. Relation between `DeltaU_(|)` and `DeltaU_(||)` cannot be determined |
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Answer» Correct Answer - C As internal energy is a point function therefore change in internal energy does not depends upon the path followed i.e., `DeltaU_(1)=DeltaU_(11)`. |
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| 796. |
A mass of dry air at NTP is compressed to (1)/(20)th of its original volume suddenly . If `gamma=1.4` , the final pressure would beA. 20 atmB. 66.28 atmC. 30 atmD. 150 atm |
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Answer» Correct Answer - B From `p_(2)V_(2)^(gamma)=p_(1)V_(1)^(gamma)` `rArr" " p_(2)=P_(1)((V_(1))/(V_(2)^(gamma)))^(gamma)=1((V_(1))/(1//20V_(1)))^(1.4)` =66.28 atm |
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| 797. |
Tyre of a bicycle has voulme `2xx10^(-3)m^(3)` Initially the tube is filled to `75%` of its volume by air at atmospheric pressure of `P_(0)=10^(5)N//m^(2)`. When a rider rides the bicycle the area of contact of tyre with road is `A=24xx10^(-5)m^(2)`. The mass of rider with bicycle is 120kg. The number of stokes which delivers, `V=500cm^(3)` volume of air in each stroke required to infalte the tyres is [Take `g=m//s^(2)`]A. 10B. 11C. 20D. 21 |
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Answer» Correct Answer - D The pressure of air in the tube is `P=(mg)/(A)+P_(0)` `P=6xx10^(5)N//m^(2)` Final volume of air in the tube is, `V=2xx10^(-3)m^(3)` The number of moles of air in the tube after pumping is, Initially volume of air in the tube is, `n=(pV)/(RT)` where T is temperature Thes n moles of air has volume `V_(1)` at atmosphere pressure given by `V_(1)=(nRT)/(P_(0))=(6xx10^(5)xx2xx10^(-3))/(10^(5))m^(3)` `=2xx10^(-3)m^(3)` Initially volume of air in the tube is `V_(0)=0.75xx2xx10^(-3)=1.5xx10^(-3)m^(3)` So, volume of air to be pumped is, `V=V_(1)-V_(0)=10.5xx10^(-5)m^(3)` So, number of strokes of pump required is, `n_(0)=(DeltaV)/(V)=(10.5xx10^(-3))/(0.5xx10^(-3))=21` |
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| 798. |
A black body at high temperature `T` `K` radiates energy at the rate of `E W//m^(2)`. When the temperature falls to `(T//2) K`, the radiated energy will beA. E/16B. E/4C. E/2D. 16 E |
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Answer» Correct Answer - A `(E_(2))/(E_(1))=((T_(2))/(T_(1)))^(4)=(((T)/(2))/(T))^(4)=(1)/(16)` `therefore E_(2)=(E)/(16)`. |
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| 799. |
A black body at a temperature of `227^(@)C` radiates heat energy at the rate of 5 cal/`cm^(2)`-sec. At a temperature of `727^(@)C`, the rate of heat radiated per unit area in cal/`cm^(2)`-sec will beA. 40B. 80C. 160D. 240 |
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Answer» Correct Answer - B `(E_(2))/(E_(1))=((T_(2))/(T_(1)))^(4)=((10000)/(500))^(4)=24=16` `E_(2)=16xx5=80` |
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| 800. |
A black body at temperature `227^(@)C` radiates heat at the rate of 5 cal/`cm^(2)`, at a temperature of `27^(@)C`, the rate of heat radiated per unit area per unit time in cal/`cm^(2)`s, isA. 4B. 1.5C. 0.64D. 0.25 |
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Answer» Correct Answer - C `(R_(2))/(R_(1))=(T_(2)^(4))/(T_(1)^(4))` `R_(2)=(81xx5)/(125)=0.64` |
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